Show that out of numbers n,n+2,n+4 only one of them is divisible by 3
Answers
theoretical explanation ..
M just explaining how because you have already got the mathematical answer..
Just imagine if the three numbers would be n, n+1 nd n+2.
It can be simply concluded that these three numbers are consecutive.
Therefore it's so obvious that one of them would definitely be divisible with 3.
Now what if we are at 3 to any of this number..?
Still we will have the same result.
ìf we add 3 to a number which is not divisible by 3 then the resulting number will also not be divisible by 3 and if we add 3 to a number which is divisible by 3 then the resulting number will also be divisible by 3.
So lets add 3 to "n+1"
We get n+4.
So now it can be concluded that any of these three number is is divisible by 3..
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We know that any positive integer of the form 3q or, 3q+1 or 3q+2 for some integer q and one and only one of these possibilities can occur.
So, we have following cases:
Case-I: When n=3q
In this case, we have
n=3q, which is divisible by 3
Now, n=3q
n+2=3q+2
n+2 leaves remainder 2 when divided by 3
Again, n=3q
n+4=3q+4=3(q+1)+1
n+4 leaves remainder 1 when divided by 3
n+4 is not divisible by 3.
Thus, n is divisible by 3 but n+2 and n+4 are not divisible by 3.
Case-II: when n=3q+1
In this case, we have
n=3q+1,
n leaves remainder 1 when divided by 3.
n is divisible by 3
Now, n=3q+1
n+2=(3q+1)+2=3(q+1)
n+2 is divisible by 3.
Again, n=3q+1
n+4=3q+1+4=3q+5=3(q+1)+2
n+4 leaves remainder 2 when divided by 3
n+4 is not divisible by 3.
Thus, n+2 is divisible by 3 but n and n+4 are not divisible by 3.
Case-III: When n=3q+2
In this case, we havE
n=3q+2
n leaves remainder 2 when divided by 3.
n is not divisible by 3.
Now, n=3q+2
n+2=3q+2+2=3(q+1)+1
n+2 leaves remainder 1 when divided by 3
n+2 is not divisible by 3.
Again, n=3q+2
n+4=3q+2+4=3(q+2)
n+4 is divisible by 3.
Hence, n+4 is divisible by 3 but n and n+2 are not divisible by 3.